I am in the process of buliding a dc motor performance testing station.In which I measure the motor current through a 10 amps 100 millivolt shunt.To convert the milli volt output from the shunt to my PLC anlog input( 0~10 Volts) I need an amplifier.Can someone help me in this regard please.

I am confused a little bit. My amplification ratio is 1:100.By various combination I can get the R2 and R1. I am an Electrical Engineer with little knowledge in Electronics.Please help.Guide me how to choose right combination with respect to input voltage and output voltage.I have a dual power supply +/- 15 ,+/- 12 volts.Can I use OP07?I am using Keyence 12 bit analog input unit PLC.

Sorry for the calculation mistake.
For 100 times R1 can be 1K and R2 can be 99 K.
The 99 K does not exist so take a 100 K.
The input can be corrected with a pot of 10 K and a resistor of 100K.
This gives you a gain range from 101 (pot at the top) and 92 (pot at the bottom.

1) Remove the signal at point A, and ground it. You will then read just the amplified offset voltage at the output of the opamp. You can use a 20k potentiometer with one end connected to pin 1, the other end connected to pin 8, and the wiper connected to V+ to adjust the offset to zero. See the datasheet for the offset nulling circuit (page 3).

2) Connect a 100mV signal to point A. Adjust R3 until you read exactly 10V out.

3) Remove the 100mV signal and ground the input. You should again read 0v at the output.

4) If you wish, connect a 50mV signal to point A. You should read 5v on the output. This will give you an indication of the linearity of the circuit.

The offset will drift over time and temperature. You will need to periodically check/readjust for this drift. As the opamp ages, it is likely that the drift will tend to stabilize somewhat; however the temperature drift will remain relatively consistent.

Wookie's circuit will work, but there is a better circuit for this application.

You really should be using a differential amplifier. This is because there is a voltage generated from the bottom of your sense resistor and ground. Secondly, as the copper heats, it changes its resistance significantly (compared to a good sense resistor).

However, attempting to build a differential amplifier out of discrete components is problematic, at the very least. Resistors have to be matched with a great deal of accuracy, or CMRR goes right out the window.

Have a look at INA118 and INA128 on Texas Instruments' website. These are reasonably inexpensive amps with nice specs that are internally trimmed; you just add a single external resistor to set the gain. A minimum of fuss to get an amp with a very high CMRR.

A buffer will be necessary if the load is of low impedance or reactive. I had simply made the assumption that your load on Vout would be high impedance (ie, DVM). With a high impedance load (Rload > 10k Ohms) the output can typically reach the supply less 2v. Since your supply is 12v, you are right at the limit.

If your load is reactive (inductive, capacitive) then you should consider using something like a Linear Technology LT1010, which is a fast 150mA power buffer. This buffer goes in the feedback loop, so there isn't an additional offset error introduced.

If you used a traditional buffer (ie: opamp with output tied to inverting input) you would need to compensate for it's offset as well.